Electronic Journal of Qualitative Theory of Differential Equations 2011, No.14, 1-18;http://www.math.u-szeged.hu/ejqtde/
Upper and Lower Solutions for BVPs on the
Half-line with Variable Coefficient and
Derivative Depending Nonlinearity
Sma¨ıl DJEBALI and Samira ZAHAR
Abstract
This paper is concerned with a second-order nonlinear boundary value problem with a derivative depending nonlinearity and posed on the positive half-line. The derivative operator is time dependent. Upon a priori estimates and under a Nagumo growth condition, the Schauder’s fixed point theorem combined with the method of upper and lower solutions on unbounded domains are used to prove existence of solutions. A uniqueness theorem is also obtained and some examples of application illustrate the obtained results.
1
Introduction
In this work, we are concerned with the existence of solutions to the following boundary value problem
x′′(t)−k2(t)x(t) +q(t)f(t, x(t), x′(t)) = 0, t >0,
x(0) = 0, x(+∞) = 0, (1.1)
whereq ∈C(0,+∞)∩L1(0,∞) while the nonlinearityf :I×R×R−→R
and the coefficient k: I →(0,∞) are continuous. Here I = (0,+∞) refers to the positive half-line.
Since BVPs on infinite intervals arise in many applications from physics, chemistry and biology, there has been so much work devoted to the inves-tigation of positive solutions for such BVPs in the last couple of years (see e.g., [2, 3, 6, 16] and the references therein) where superlinear or sublin-ear nonlinsublin-earities are considered. The positivity of solutions is motivated by the fact that the unknown x may refer to a density, a temperature or the concentration of a product. For instance, the linear operator of deriva-tion −x′′+cx′+λx(c, λ >0), which may be rewritten in reduced form as
2000 Mathematics Subject Classification. 34B15, 34B18, 34B40.
−x′′+k2x,stems from epidemiology and combustion theory and models the propagation of the wave front of a reaction-diffusion equation (see e.g., [6]). Methods used to investigate these problems range from the upper and lower solution techniques [15, 17] to the fixed point theory in weighted Banach spaces and the index fixed point theory on cones of some Banach spaces [1, 5, 16, 18].
When k is constant, the existence of solutions to problem (1.1) was established in [14] using the Tychonoff’s fixed point theorem. It was also studied by Djebaliet al in [7, 8, 9] where multiplicity results have been also given. In [17], B. Yanet alhave used the upper and lower solution techniques to obtain some existence results whenf is allowed to have a singularity at
x= 0 and may change sign.
In the general case when the constantkis replaced by a bounded function
k=k(t),the problem was recently investigated by Ma and Zhu in [13]. The nonlinearityf ∈C(R+×R+,R) is assumed to satisfy a sublinear polynomial growth condition. The authors of [13] proved that if the parameterλis less that someλ0,then the following problem
x′′(t)−k2(t)x(t) +λq(t)f(t, x(t)) = 0, t >0,
x(0) = 0, x(+∞) = 0,
has a positive solution; a fixed point theorem in a cone of a Banach space has been employed. Their investigation relies heavily on estimates of the corresponding Green’s function. In [11], the authors first applied fixed point index theory in cones of Banach spaces to prove existence results when
f = f(t, x) is positive and may exhibit a singularity at the origin with respect to the solution; then they used the Schauder’s fixed point theorem together with the method of upper and lower solutions to prove existence of solutions whenf is not necessarily positive.
Using an upper and lower solution method on infinity intervals, the aim of this paper is to investigate the more general problem where the non-linearity f = f(t, x, y) is derivative depending. In [12], Lian et al. used unbounded upper and lower solutions on noncompact intervals to prove an existence result for a class of BVPs. In [10], the authors considered prob-lem (1.1) withq= 1 and used topological degree theory combined with the existence ofCB1 upper and lower solutions to prove existence of solutions on bounded intervals. Solutions are then extended to the positive half-line by means of sequential arguments. In the present paper, we complement these existence theorems via a direct approach.
on suitable a priori estimates. A uniqueness result is also given in Section 5. Finally, we give two examples of application to illustrate our existence and uniqueness results.
2
Auxiliary Lemmas
Let us first enunciate an assumption regarding the functionk: (H0) the function k:I →[0,∞) is bounded and continuous and
∃d∈[k, k], ∀ρ >0, lim t→∞e
−ρtZ t
0
eρs[k2(s)−d2]dsexists
where k:= sup t∈[0,∞)
k(t) and k:= inf
t∈[0,∞)k(t)>0.
In order to construct a Green’s function of the corresponding linear problem, it is necessary to know a fundamental system of solutions. The following auxiliary results are brought from [13].
Lemma 2.1. Assume that k is bounded and continuous. Then the Cauchy
problem
x′′(t)−k2(t)x(t) = 0, t >0,
x(0) = 0, x′(0) = 1 (2.1)
has a unique solution φ1 defined on [0,+∞). Moreover φ1 is nondecreasing
and unbounded.
Lemma 2.2. (See also [1], Thm. 7) Assume that k is bounded and contin-uous. Then the problem
x′′(t)−k2(t)x(t) = 0, t >0,
x(0) = 1, limt→+∞x(t) = 0 (2.2)
has a unique solution φ2 defined on [0,+∞) with
0< φ2 ≤1, φ′2 <0.
If further (H0) holds, then
lim t→∞
φ′2(t)
φ2(t)
=−d.
Lemma 2.3. Assume that (H0) holds. Then there exists M >0 such that
supt∈[0,∞)φ1(t)φ2(t)< M.
Lemma 2.4. Assume (H0) holds. Then for any functiony ∈L1[0,∞),the
problem
x′′(t)−k2(t)x(t) +y(t) = 0, t >0,
is equivalent to the integral equation
x(t) =
Z ∞
0
G(t, s)y(s)ds, t >0,
where
G(t, s) =
φ1(t)φ2(s), s≥t,
φ1(s)φ2(t), t≥s. (2.3)
Lemma 2.5. For all (t, s)∈[0,∞)×[0,∞), G(t, s)< 21k·
Moreover, we have
Lemma 2.6. For any t∈(0,∞), we have
0≤φ′1(t)φ2(t)≤1
and
−1≤φ1(t)φ′2(t)≤0.
Proof. Since {φ1, φ2}is the fundamental system, we have that φ1(t)φ′2(t)−
φ′1(t)φ2(t) =−1 which means that φ′1(t)φ2(t) + [−φ1(t)φ′2(t)] = 1.Then our
claim follows from the sign and the monotonicity ofφ1, φ2.
Lemma 2.7. We have
k0:=
Z ∞
0
k2(s)φ2(s)ds <∞.
Proof. By Lemma 2.2, it is clear that the function U = φ
′
2
φ2 satisfies the Ricatti equation U′+U2 =k2 both with the terminal conditionU(+∞) =
−d. Hence if, by contradiction, U(0) = −∞, then U′(0) = −∞, which is impossible. Thus −∞< U(0)<0 and as a consequence −∞ < φ′2(0)<0 which implies that 0< k0 =−φ′2(0)<∞,as claimed.
Now we define what we mean by lower and upper solutions.
Definition 2.1.
(a) We say that α is a lower solution of problem (1.1) if α ∈ C1[0,∞) ∩
C2(0,∞) and
α′′(t)−k2(t)α(t) +q(t)f(t, α(t), α′(t))≥0, α(0)≤0, α(+∞)≤0.
(b) A function β is an upper solution of problem (1.1) if β ∈ C1[0,∞) ∩
C2(0,∞) and
3
General Assumptions and a Modified Problem
We first posit some assumptions:
(H1) There exist α≤β lower and upper solutions of problem (1.1)
respec-tively.
(H2) α0 := sup
t∈[0,∞)
{|α(t)|φ−21(t)}<∞ and β0 := sup
t∈[0,∞)
{|β(t)|φ2−1(t)}< ∞,
whereφ−21(t) := φ21(t).
(H3) There exist continuous functions ψ : I → [0,∞) and h :R → [1,∞)
such that Z ∞
0
ψ(s)q(s)ds <∞, (3.1)
Z ∞
0
ds
h(s) = +∞ (3.2)
and
|f(t, x, y)| ≤ψ(t)h(y), ∀(t, x, y)∈Dβα×R, (3.3)
where Dαβ is defined by
Dβα:={(t, x)∈(0,∞)×R:α(t)≤x≤β(t)}.
(H4) α1 := supt∈R+α′(t) < ∞ , β1 := inft∈R+β′(t) > −∞, and for any
y∈R andt∈(0,∞), we have
y < β′(t) =⇒ f(t, β(t), y)≤f(t, β(t), β′(t))
y > α′(t) =⇒ f(t, α(t), y)≥f(t, α(t), α′(t)).
(H4)′
α1 := sup
t∈R+
|α′(t)|<∞ and β1 := sup
t∈R+
|β′(t)|<∞.
Now, define the Banach space
X={x∈ C1[0,∞) : lim
t→+∞x(t) and t→lim+∞x
′(t) exist}
equipped with the normkxk= max
(
sup t∈[0,∞)
|x(t)|, sup t∈[0,∞)
|x′(t)|
)
.
The following compactness criterion will be needed (see [4], p. 62).
Lemma 3.1. Let M ⊆X. Then M is relatively compact in X if the fol-lowing conditions hold
(b) the functions belonging to M and the functions belonging to {u: u(t) =
x′(t), x∈M} are locally equicontinuous on[0,+∞),
(c) the functions belonging to M and the functions belonging to {u: u(t) =
x′(t), x∈M} are equiconvergent at+∞.
Given two continuous functions α andβ such thatα≤β, we define the truncated functionfeby
e
f(t, x, y) =
fR(t, β(t), y) +1+β|β(t()t−)−xx|, β(t)< x,
fR(t, x, y), α(t)≤x≤β(t),
fR(t, α(t), y) + 1+x|α−α(t()t−)x|, x < α(t).
where
fR(t, x, y) =
f(t, x,−R), y <−R, f(t, x, y), |y| ≤R, f(t, x, R), R < y
and the real number R is such that R > max{|α1|,|β1|}. Finally, consider
the modified problem
x′′(t)−k2(t)x(t) +q(t)fe(t, x(t), x′(t)) = 0, t∈(0,∞)
x(0) = 0, x(+∞) = 0. (3.4)
4
Existence Results
4.1 A priori Estimates
Proposition 4.1. Assume that either (H1) and (H4), or (H1) and (H4)′
hold. Then all possible solutions of problem (3.4) satisfy
α(t)≤x(t)≤β(t), ∀t∈I.
Proof. We prove that x(t) ≤ β(t),∀t ∈ I. Suppose, on the contrary that
supt∈[0,∞)(x−β)(t) > 0. Since (x−β)(+∞) = −β(+∞) ≤ 0 and (x−
β)(0) =−β(0)≤0,then there exists t0 ∈(0,∞) such that x(t0)−β(t0) =
sup(x−β)(t0)>0; hence (x′′−β′′)(t0)≤0 andx′(t0)−β′(t0) = 0. Moreover,
by definition of an upper solution, we have the successive estimates:
(x′′−β′′)(t
0) = k2(t0)x(t0)−q(t0)fe(t0, x(t0), x′(t0))−β′′(t0)
≥ k2(t0)x(t0)−q(t0)fe(t0, x(t0), x′(t0))−k2(t0)β(t0)
Hence
(x′′−β′′)(t0) ≥ k2(t0)(x−β)(t0)−q(t0)fe(t0, x(t0), β′(t0))
+q(t0)f(t0, β(t0), β′(t0))
= k2(t0)(x−β)(t0)−q(t0)fR(t0, β(t0), β′(t0))
−q(t0)1+(β|(−β−x)(x)(t0t0))|+q(t0)f(t0, β(t0), β′(t0))
> −q(t0) [fR(t0, β(t0), β′(t0))−f(t0, β(t0), β′(t0))].
To check that the last right-hand term is nonnegative, we distinguish be-tween two cases:
(a) In case (H4) holds, consider the sub-cases:
(a1) β′(t
0)<−R implies that |β1|> R which does not hold true.
(a2) If −R≤β′(t0)≤R, thenfR(t0, β(t0), β′(t0)) =f(t0, β(t0), β′(t0)).
(a3) Ifβ′(t0)> R, thenfR(t0, β(t0), β′(t0)) =f(t0, β(t0), R)≤f(t0, β(t0), β′(t0))
follows from the first part of (H4).
(b) If (H4)′holds then−R ≤β′(t0)≤R.ConsequentlyfR(t0, β(t0), β′(t0)) =
f(t0, β(t0), β′(t0)).Our claim is then proved leading to a contradiction.
Similarly, we can prove that x(t)≥α(t) for everyt∈[0,∞).
Remark 4.1. Assumption (H4) is essential in Proposition 4.1. Such an
hypothesis is missing to complete the proof of Theorem 3.1 in [12].
4.2 The Truncated Problem
Theorem 4.1. Under Assumptions (H0), (H1), and (H3), the truncated
problem (3.4) has at least one solution inX.
Proof. Since solving problem (3.4) amounts to proving existence of a fixed
point forT, let us consider the operator T :X −→X defined by
(T x)(t) =
Z ∞
0
G(t, s)q(s)fe(s, x(s), x′(s))ds. (4.1)
(a) T :X−→X is well defined. Let x∈X. From (3.1) and (3.3), we get
(T x)(t) ≤ R0∞G(t, s)q(s)fe(s, x(s), x′(s))ds
≤ R0∞G(t, s)q(s) (ψ(s)h(x′(s)) + 1)ds
where H0 = H0(x) = max 0≤t≤||x′
||h(t). From the monotonicity of φ1 and φ ′
1
together with Lemma 2.6, we obtain that
(b) T :X −→ X is continuous. Let (xn)n∈N be a sequence converging to
Now, givenT >0 andt0, t1 ∈[0, T], we have the estimates
|(T x)(t0)−(T x)(t1)| ≤
R∞
0 |G(t0, s)−G(t1, s)|q(s)(Hrψ(s) + 1)ds
≤ R0T |G(t0, s)−G(t1, s)|q(s)(Hrψ(s) + 1)ds
+RT∞|φ1(t0)φ2(s)−φ1(t1)φ2(s)|q(s)(Hrψ(s) + 1)ds
≤ R0T |G(t0, s)−G(t1, s)|q(s)(Hrψ(s) + 1)ds
+|φ1(t0)−φ1(t1)|
R∞
0 q(s)(Hrψ(s) + 1)ds.
By (3.1), the continuity of the Green’s function and the Lebesgue dominated convergence theorem, we get
lim |t1−t0|→0
Z T
0
|G(t0, s)−G(t1, s)|q(s)(Hrψ(s) + 1)ds= 0.
In addition, (3.1) and the continuity ofφ1 imply that
lim
|t1−t0|→0|φ1(t0)−φ1(t1)|
Z ∞
0
q(s)(Hrψ(s) + 1)ds= 0.
Hence the right-hand term goes to 0 as |t1−t0| →0. Moreover, fort0 ≤t1,
the following estimates hold true
|(T x)′(t0)−(T x)′(t1)| ≤
R∞
0 |Gt(t0, s)−Gt(t1, s)|q(s)(Hrψ(s) + 1)ds
= R0t0|Gt(t0, s)−Gt(t1, s)|q(s)(Hrψ(s) + 1)ds
+Rt0t1|Gt(t0, s)−Gt(t1, s)|q(s)(Hrψ(s) + 1)ds
+Rt1T |Gt(t0, s)−Gt(t1, s)|q(s)(Hrψ(s) + 1)ds
+RT∞|Gt(t0, s)−Gt(t1, s)|q(s)(Hrψ(s) + 1)ds
≤ R0t0φ1(s)|φ′2(t0)−φ′2(t1)|q(s)(Hrψ(s) + 1)ds
+Rt0t1|φ′
1(t0)φ2(s)−φ1(s)φ′2(t1)|q(s)(Hrψ(s) + 1)ds
+Rt1T φ2(s)|φ′1(t0)−φ′1(t1)|q(s)(Hrψ(s) + 1)ds
Consequently,
|(T x)′(t0)−(T x)′(t1)| ≤ φ1(t0)|φ′2(t0)−φ′2(t1)|
Rt0
0 q(s)(Hrψ(s) + 1)ds
+(|φ′
1(t1)φ2(t1)|+|φ1(t1)φ′2(t1)|)Rt0t1q(s)(Hrψ(s) + 1)ds
+|φ′1(t0)−φ′1(t1)|Rt1T q(s)(Hrψ(s) + 1)ds
+|φ′1(t0)−φ′1(t1)|RT∞q(s)(Hrψ(s) + 1)ds,
each of the four terms above tends to 0 as|t1−t0|tends to 0, proving that
T xis almost equicontinuous.
To prove equiconvergence, we first notice that limt→∞T x(t) = 0. More-over from lim
t→∞φ2(t) = 0 and
R∞
0 q(s)(Hrψ(s) + 1)ds < ∞, for any ε > 0,
there exists N >0 such that for t≥N,the following estimates hold true:
0 ≤ sup
x∈B
|T x(t)−0| ≤sup x∈B
R∞
0 G(t, s)q(s)|fe(s, x(s), x′(s))|ds
≤ R0tφ1(s)φ2(t)q(s)(Hrψ(s) + 1)ds+
R∞
t φ1(t)φ2(s)q(s)(Hrψ(s) + 1)ds = R0Nφ1(s)φ2(t)q(s)(Hrψ(s) + 1)ds+
Rt
Nφ1(s)φ2(t)q(s)(Hrψ(s) + 1)ds +Rt∞φ1(t)φ2(s)q(s)(Hrψ(s) + 1)ds
≤ R0Nφ1(s)φ2(t)q(s)(Hrψ(s) + 1)ds+
R∞
N φ1(t)φ2(t)q(s)(Hrψ(s) + 1)ds +RN∞φ1(t)φ2(t)q(s)(Hrψ(s) + 1)ds
≤ φ1(N)φ2(t)R0Nq(s)(Hrψ(s) + 1)ds+M
R∞
N q(s)(Hrψ(s) + 1)ds
+MRN∞q(s)(Hrψ(s) + 1)ds.
Hence
sup x∈B
|T x(t)| ≤ ε
3 +
ε
3 +
ε
3 =ε.
Furthermore, for anyε >0, there existsN >0 such that for t≥N
|φ1(s)φ′2(t)| = |G(t, s)(
φ′
2(t)
φ2(t)+d−d)|
≤ G(t, s)|φ
′
2(t)
φ2(t)+d|+dG(t, s)
≤ 2R∞ ε
and
As a consequence, for t≥N, we obtain the estimates
sup
Thus we have proved equiconvergence ofT ending the proof that T is com-pletely continuous. Finally, by the Leray-Schauder fixed point theorem, we deduce thatT has at least a fixed pointx, solution of problem (3.4).
(a) By Theorem 4.1, problem (3.4) has at least one solution inX. In addition, proposition 4.1 implies that any solution x of problem (3.4) satisfies the bounds
For the sake of brevity, we only consider the first case. Using the fact that
Then x′(t1)≤R. Since t0 and t1 are arbitrary, we obtain that if x′(t) ≥η,
then x′(t) ≤R, t ∈[0,∞) yielding that fR(t, x(t), x′(t)) = f(t, x(t), x′(t)). This means thatx is a solution of problem (1.1), which completes the proof of the theorem.
Remark 4.2. The condition h(s) ≥ 1 in (H3) is not essential; in fact it
is sufficient to suppose h(s) ≥ h0 for some h0 > 0. Indeed, in this case
h(s)
h0 ≥1 and then we have to write in the above estimates:
Rx′
(t1)
x′
(t0)
ds h(s) ≤
Z t1
t0
k2(s)|x(s)|+q(s)ψ(s)h(x′(s))
h(x′(s)) ds
≤
Z t1
t0
h(x′(s))( 1
h0k2(s)|x(s)|+q(s)ψ(s))
h(x′(s)) ds
≤ Rt0t1 h01 k2(s)|x(s)|+Rt0t1q(s)ψ(s))ds
≤ h01 k0max(β0, α0) +R0∞q(s)ψ(s))ds.
So we have just to modify (4.3) by
Z R
η
ds h(s) >
1
h0
k0max(β0, α0) +
Z ∞
0
q(s)ψ(s))ds.
Our second existence result is
Theorem 4.3. Assume that all conditions of Theorem 4.2 are satisfied but
(H2) replaced by
(H2)′ k1 := R0∞k2(t) max(|α(t)|,|β(t)|)dt < ∞. Then problem (1.1) has at
least one solution x having the representation
x(t) =
Z ∞
0
G(t, s)q(t)f(s, x(s), x′(s))ds
and such that
α(t)≤x(t)≤β(t), t∈[0,∞).
Proof. From (3.2), we can find real numbersR >e max{|α1|,|β1|}and η >0
such that
Z Re
η
ds
h(s) ≥k1+
Z ∞
0
ψ(s)q(s)ds. (4.4)
Then the proof runs parallel to the proof of Theorem 4.2 withRreplaced by
e
estimates instead:
Rx′
(t1)
x′
(t0)
ds h(s) =
Rt1 t0
x′′
(s)
h(x′
(s))ds=
Rt1 t0
k2(s)x(s)−q(s)f
R(s,x(s),x′(s))
h(x′
(s)) ds
≤ Rt0t1 k2(s)|x(s)|h+(qx(′s(s)ψ))(s)h(x′(s))ds
≤ Rt0t1 h(x′(s))(k2h(s()x|x′((ss)))|+q(s)ψ(s))ds
≤ Rt0t1k2(s)|x(s)|+Rt1
t0 q(s)ψ(s)ds
≤ k1+
R∞
0 q(s)ψ(s)ds
≤ RηRe hds(s).
Finally, we complete the proof using (4.4).
Remark 4.3. Contrarily to (H2), assumption (H2)′ allows the upper and
lower solutions to be unbounded.
5
A Uniqueness Result
The following result complements Theorems 4.2 and 4.3.
Theorem 5.1. Assume that f =f(t, x, y) is continuously differentiable in
x andy for eacht≥0 and satisfies either the conditions of Theorem 4.2 or
Theorem 4.3 together with
(H5) f(t, x, y) is nonincreasing inx for each t and y fixed.
Then problem (1.1) has a unique solution x such that
α(t)≤x(t)≤β(t), ∀t≥0.
Proof. Suppose there exist two distinct solutionsx1,x2of problem (1.1) and
letz:=x1−x2. By the mean value theorem, there existθ, ϕ such that
f(t, x2, x′2) =f(t, x1, x′1)−z
∂f
∂x(t, θ, ϕ)−z
′∂f
∂y(t, θ, ϕ).
Assume that z(t1) > 0 for some t1 and that z has a positive maximum at
somet0<∞. Then, with (H5), we have
0≥z′′(t0) = k2(t0)z(t0) +q(t0) [f(t0, x2(t0), x′2(t0))−f(t0, x1(t0), x′1(t0))]
= k2(t0)z(t0)−q(t0)∂f∂x(t0, θ, ϕ))z(t0)−q(t0) ∂f
∂y(t0, θ, ϕ)z′(t0)
= z(t0)
h
k2(t0)−q(t0)∂f∂x(t0, θ, ϕ)
leading to a contradiction. Hence supz(t) = limt→∞z(t). But limt→∞z(t) = limt→∞[x1(t)−x2(t)] = 0, which is again a contradiction, ending the proof
of the theorem.
6
Applications
6.1 Example 1
Consider the boundary value problem
x′′(t)−k2(t)x(t) +q(t)[x(t)(x′)θ(t) +c(t)] = 0, t >0,
x(0) = 0, x(+∞) = 0, (6.1)
whereθ = 2p1+1 and p is a positive integer. The positive functions c=c(t) and q = q(t) satisfy q(t), φ2(t)q(t) ∈ C(0,+∞)∩L1(0,∞) and 0 ≤ c(t) ≤
−φ2(t)(φ′2)θ(t).The function kverifies (H0).
Thenα(t)≡0 andβ(t) =φ2(t) are respectively lower solution and upper
solution withα≤β.Moreover
α0 = sup
t∈[0,∞)
{|α(t)|φ2−1(t)}= 0 and β0= sup
t∈[0,∞)
{|β(t)|φ−21(t)}= 1.
Then (H1) and (H2) are satisfied. As for (H3), one may take ψ(t) = φ2(t)
andh(y) = (|y|+ 1 +γ)≥1 withγ :=−(φ′2)θ(0) so that, for 0≤x≤φ
2(t),
we have
|f(t, x, y)| =|xyθ+c(t)| ≤φ
2(t)|y|θ−φ2(t)(φ′2)θ(t)
≤φ2(t)(|y|θ−(φ′2)θ(0))≤φ2(t)(|y|+ 1 +γ)
as well as
Z ∞
0
ψ(s)q(s)ds <∞, and
Z ∞
0
ds h(s) =
Z ∞
0
ds
s+ 1 +γ = +∞.
Regarding (H4),we have
α1 := supt∈R+α′(t) = 0
β1:= inft∈R+φ′
2(t) =φ′2(0) =−
R∞
0 φ′′2(s)ds=−
R∞
0 k2(s)φ2(s)ds=−k0.
In addition, for anyy∈Rand t∈(0,∞),we have
y < β′(t) =⇒f(t, β(t), y) =φ2(t)yθ+c(t)
≤φ2(t)φ′2θ(t) +c(t) =f(t, β(t), β′(t))
y > α′(t) =⇒y >0 =⇒ f(t, α(t), y) =c(t) =f(t, α(t), α′(t)).
Therefore, Theorem 4.2 yields that problem (6.1) has at least one solution
x such that
6.2 Example 2
Further to (H1)−(H4), it is clear that the nonlinearityf(t, x, y) =−exe−y+
c(t) satisfies (H5). Arguing as in Example 1, we can see that Theorems 4.2
and 5.1 imply that the boundary value problem
x′′(t)−k2(t)x(t) +q(t)[−ex(t)e−x′
(t)+c(t)] = 0, t >0,
x(0) =x(+∞) = 0,
where the positive functionsc = c(t) and q = q(t) satisfy q(t), φ2(t)q(t) ∈
C(0,+∞)∩L1(0,∞) and 1 ≤ c(t) ≤ eφ2(t)e−φ′
2(t), t ≥ 0, has exactly one solution xsatisfying
06x(t)6φ2(t), ∀t≥0.
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Sma¨ıl DJEBALI (e-mail: [email protected])
Department of Mathematics, E.N.S. P.B. 92 Kouba, 16050. Algiers, ALGERIA
Samira ZAHAR (e-mail: zahar [email protected])
Department of Mathematics, University Abd-Errahmane Mira, 06000. Bejaia, Algeria
